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Alekssandra [29.7K]
4 years ago
7

According to the concept of length contraction, what happens to the length of an object as it approaches the speed of light and

then slows down, eventually coming to a stop? It is always observed as contracting. It is always observed as the same length. It is observed as expanding and then contracting back to its original length. It is observed as contracting and then expanding back to its original length.
Physics
1 answer:
lord [1]4 years ago
8 0
When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.
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Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

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Vf₁ = Final Speed of Car = ?

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a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

<u>Vf₁ = 25 m/s</u>

<u></u>

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

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Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

<u>s₂ = 90 m</u>

<u></u>

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

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For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

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Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

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Answer:

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Explanation:

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